Find the value of $x$ if $x$ is positive and $x\cdot\lfloor x\rfloor=70$. Express your answer as a decimal.
Solution: We know that $\lfloor x\rfloor \leq x < \lfloor x\rfloor + 1$. This implies that $\lfloor x\rfloor^2 \leq x\cdot\lfloor x\rfloor < \left(\lfloor x\rfloor + 1\right)^2$ for all values of $x$. In particular since $x\cdot\lfloor x\rfloor=70$ and $8^2<70<9^2$, we can conclude that $8<x<9\Longrightarrow\lfloor x\rfloor=8$. From there, all we have to do is divide to get that $x=\frac{70}{8}=\boxed{8.75}$.